GATE CS 2014 Set 1 QGA-10 — General Aptitude

Let the four vertices of the tetrahedron be $A, B, C,$ and $D$. Let the point inside the tetrahedron be $P$.

Connecting $P$ to each of the four corners gives four new line segments: $PA, PB, PC,$ and $PD$.

An internal plane is formed by any two of these intersecting lines. Since all four lines meet at the common point $P$, any pair of these lines defines a unique plane.

The number of ways to choose $2$ lines out of the $4$ lines to form a plane is given by the combination:

These $6$ planes are:

• Plane $PAB$
• Plane $PAC$
• Plane $PAD$
• Plane $PBC$
• Plane $PBD$
• Plane $PCD$

Since $P$ lies strictly inside the tetrahedron, none of these planes coincide with each other or with the original four boundary faces ($ABC, ABD, ACD, BCD$). Thus, exactly 6 new internal planes are created.